JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    If the line, \[\frac{x-3}{1}=\frac{y+2}{1}=\frac{z+\lambda }{-2}\] lies in the plane, \[2x-4y+3z=2,\] then the shortest distance between this line and the line, \[\frac{x-1}{12}=\frac{y}{9}=\frac{z}{4}\] is:      [JEE Online 09-04-2017]

    A)  1                                            

    B)  2

    C)  3                                            

    D)  0

    Correct Answer: B

    Solution :

      \[pt\,(3,\,-2,\,\lambda )\] on pline \[2x-4y+3z-2=0\] \[=6+8-3\lambda -2=0\] \[=3\lambda =12\,\,\] Now \[\frac{x-3}{1}=\frac{y+2}{-1}\,=\frac{z+4}{-2}={{k}_{1}}\]                            ?(1) \[\frac{x-1}{12}=\frac{y}{9}=\frac{z}{4}\,={{k}_{2}}\]                                       ?(2) pt on given \[1=({{k}_{1}}+3,\,-{{k}_{1}}-2,\,-2{{k}_{1}}-4)\] \[pt\] on given \[2=\,(12\,{{k}_{2}}\,+1,\,\,9{{k}_{2}},\,4{{k}_{2}})\] \[{{k}_{1}}+3=12{{k}_{2}}+1|-{{k}_{1}}-2=9{{k}_{2}}|-2{{k}_{1}}-4=4{{k}_{2}}\] \[\left. \begin{align}   & {{k}_{2}}=0 \\  & {{k}_{1}}=-2 \\ \end{align} \right]\] \[p\,(1,\,\,0,\,\,0)\] gives are ditersech - thortest distance = 0


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